CAT 2008: Quantitative Questions a Day 134 -Till end -> The Discussions

shivam_01

Complex problem to start with....

take the series of 20 and try to generalise it
1234567891011121314151617181920

after first elimination
series left is
23568911121415171820
Numbers cancelled of the order = 3K+1

second elimination
series left is
358912141718
Numbers cancelled of the order = 4K'+2 or 9k''+2 or 13k'''+2....

Third elimination
series left is
58121418
Numbers cancelled of the order = 6K'''+3 / 14k''''+3

Fourth elimination
series left is
81218
Numbers cancelled of the order = 9K'''''+5

Fifth elimination
series left is
1218
Numbers cancelled of the order = 13n+8

Sixth elimination
series left is
18
Numbers cancelled of the order = 12x+18

Now checking the answer choices

140 does not satisfies Option (e)

I doubt whether this is completely true and also in case one option is none of these .. then what ?

daze, amit and implex bhai logo kya program likha hain .... awesome great
hats off to u

Aaarav ye official solution mein mat dena plzzz...

nbangalorekar

I think the best way to solve this problem wud be to use recursion.... however im still not able to formulate a soln in my mind!!

nbangalorekar

Quote:

Originally Posted by nbangalorekar

I think the best way to solve this problem wud be to use recursion.... however im still not able to formulate a soln in my mind!!

one thing is clear that the (3k+2) position will become (2k+1) & 3k position will become (2k)... so we can write:
f(3k+2)=2k+1
f(3k)=2k
we will have to use recurssion on this...

nbangalorekar · 12:49 PM

Quote:

Originally Posted by nbangalorekar

one thing is clear that the (3k+2) position will become (2k+1) & 3k position will become (2k)... so we can write:
f(3k+2)=2k+1
f(3k)=2k
we will have to use recurssion on this...

Aarav kuch to help karo after this yaar!! even my work is held up.... not able to concentrate

it is frustrating!!
EDIT: work had muddled my mind....got the answer finally (posted belw!!)

vineet.nitd · 11:45 AM

We can solve this by back trcing...
After 2 steps number of terms reduced are 50 and 34 respectively, with end pointer pointing at 99 on 66th position.

After the 1st step, 140(93),141(94),143(95)144(96),146(97),147(9

,149 (99),150(100)

After step 2,
93(62), 95(63), 96(64), 98(65), 99(66)
Numbers in brackets are the new position of the numbers.
One can iterate further but it is easier to find the value for the first 66 numbers which comes out to be 62.

The number associated with the position 62 is 93 which in turn is the position of 140 after the 1st step. Hence 140 is the right ans..

v-factor

My approach:

In the 1st round all nos eliminated are of form 3n+1(1,4,7,10....)

In 2nd round nos. eliminated are (2,6,11,15,20......)
We see that the pattern of differences is 4,5 alternately
So nos. eliminated in this round are of form 4n+2 or 9n+2
38 = 9*4+2
Option D) gone

In 3rd round nos. eliminated are (3,9,17,23......)
We see that the pattern of differences is 6,8 alternately
So nos. eliminated in this round are of form 6n+3 or 14n+3 or 14n-5
93 = 14*7-5
Option A) gone

In 4th round nos. eliminated are (5,14,26......)
We see that the pattern of differences is 9,12 alternately
So nos. eliminated in this round are of form 9n+5 or 21n+5 or 21n-7
119 = 21*6 - 7
Option C) gone

In 5th round nos. eliminated are (8,21,39......)
We see that the pattern of difference is 13,18 alternately
So nos. eliminated in this round are of form 13n+8 or 31n+8 or 31n-10

In 6th round we get eliminated nos. as 18n+12
48=18*2+12
Option B0 gone

only E)140 remains

singh_kanwar

hey guys with due concern to ur hard work in solving such a hectic question of today i sticked wid the old approach of problem solving .......MANUAL CALCULATION
i knw its a hectic n time consuming job bt i m quite a lazy wen it comes to solving via algorithms.....so cudnt resist myself n choosen d best among d wrost options to solve d question.......and the answer is (e) 140

p.s. - please help me guys in quant. i badly screwed up d last aimcat of time n its really disgusting to score such unbearable marks jst a month b4 CAT......plz can somebody refer how sud i manage it???i mean frm where sud i practice quant?i ve finished d material provided by time n it seems to b of no help

shivam_01

Quote:

Originally Posted by singh_kanwar

hey guys with due concern to ur hard work in solving such a hectic question of today i sticked wid the old approach of problem solving .......MANUAL CALCULATION
i knw its a hectic n time consuming job bt i m quite a lazy wen it comes to solving via algorithms.....so cudnt resist myself n choosen d best among d wrost options to solve d question.......and the answer is (e) 140

p.s. - please help me guys in quant. i badly screwed up d last aimcat of time n its really disgusting to score such unbearable marks jst a month b4 CAT......plz can somebody refer how sud i manage it???i mean frm where sud i practice quant?i ve finished d material provided by time n it seems to b of no help

sorry for spamming here .. but kanwar in case u have finished the time material completely i see no point that u are unable to solve teh questions ..

Time material is huge and when i see to it i see no point to complete it by any means... I guessed u solved it in a hurry . Please revise the same and practice mocks

nbangalorekar

Quote:

Originally Posted by vineet.nitd

We can solve this by back trcing...
After 2 steps number of terms reduced are 50 and 34 respectively, with end pointer pointing at 99 on 66th position.

After the 1st step, 140(93),141(94),143(95)144(96),146(97),147(9

,149 (99),150(100)

After step 2,
93(62), 95(63), 96(64), 98(65), 99(66)
Numbers in brackets are the new position of the numbers.
One can iterate further but it is easier to find the value for the first 66 numbers which comes out to be 62.

The number associated with the position 62 is 93 which in turn is the position of 140 after the 1st step. Hence 140 is the right ans..

Vineet sir.... we can also solve tis qn even by a str8 forward approach...(struck me later)...
in the end only 2 ppl will remain.
so obviously the 2nd person has to win..
as is obvious using recursion (3k+2) will become 2k+1 & 3k will become 2k
now 2 is of the form 2k. so it wud have been 3k (ie 3)
this way just use recursion, the answer will come to 140!!

nbangalorekar

Quote:

Originally Posted by nbangalorekar

Vineet sir.... we can also solve tis qn even by a str8 forward approach...(struck me later)...
in the end only 2 ppl will remain.
so obviously the 2nd person has to win..
as is obvious using recursion (3k+2) will become 2k+1 & 3k will become 2k
now 2 is of the form 2k. so it wud have been 3k (ie 3)
this way just use recursion, the answer will come to 140!!

one thing iv realised is the best method to solve such questions is to find the recursive behavior...